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• Level: GCSE
• Subject: Maths
• Word count: 7068

Extracts from this document...

Introduction

Aim: To investigate the gradient of different curves

I will investigate the gradient of different equations for which the general formula is

Y = aχn

In this equation I will investigate the gradient by varying the values of ‘a’ and ‘n’.

The property possessed by a line or surface that departs from the horizontal is called the gradient of the line. In mathematical terms, the gradient of the line simply tells us how steep a line is. The gradient for all lines parallel to the X-axis is 0.

Gradient function is the name of a rule, specific to a graph ex: Y = X3 which can be used to find the gradient at any point of the graph. The ‘X’ value is substituted in the equation and this gives the exact gradient for that specific graph.

I plan to find the gradient using these two methods:

Tangent Method:

The tangent method involves making a tangent at a point on the graph on which the gradient is to be found. The figure below shows a tangent:

To find the gradient of a straight line, we use the equation:

Y2-Y1

X2-X1

This equation would give us the gradient. [X1, Y1], [X2, Y2] can be any two points on the graph. This equation can be used only on straight-line graphs. To find the gradient of curves is much more difficult than the straight-line graphs. For the gradient of curves, a tangent is drawn at a point. A tangent is a line, which touches the graph at one point and one point only. To find the gradient of the curve at the specific point, we then find the gradient of the tangent using the formula given above.

This method is quite inaccurate when done manually, but it is the easiest.

Middle

-0.33344448

(X1, Y1) = -3, -1

(X2, Y2) = -2.999, (3/X2)

(X2, Y2) = (-2.999, -1.00033)

= -1.00033-(-1)/-2.999-(-3)

= -0.33344448

Graph: Y = -2/x

The table below shows the increment method performed, at the point (-2, 1)

 Y =-2/x X1 Y1 X2 Y2 Gradient -2 1 -1.995 1.002506 0.50125313 -2 1 -1.996 1.002004 0.501002 -2 1 -1.997 1.001502 0.50075113 -2 1 -1.998 1.001001 0.5005005 -2 1 -1.999 1.0005 0.50025013

(X1, Y1) = -2, -1

(X2, Y2) = -1.999, (-2/X2)

(X2, Y2) = (-1.999, 1.0005)

= 1.0005-1/-1.999-(-2)

= 0.50025013

The table below shows the increment method performed at the point (-3.5, 0.57)

 Y =-2/x X1 Y1 X2 Y2 Gradient -3.5 0.57 -3 0.667 0.19 -3.5 0.57 -3.1 0.645 0.19 -3.5 0.57 -3.2 0.625 0.18 -3.5 0.57 -3.3 0.606 0.18 -3.5 0.57 -3.4 0.588 0.18

(X1, Y1) = -3.5, 0.57

(X2, Y2) = -3.4, (-2/X2)

(X2, Y2) = (-3.4, 0.588)

=  0.588-0.57/-3.4-(-3.5)

= 0.18

The table below shows the increment method performed at the point (2, -1)

 Y =-2/x X1 Y1 X2 Y2 Gradient 2 -1 2.005 -0.997506 0.49875312 2 -1 2.004 -0.998004 0.499002 2 -1 2.003 -0.998502 0.49925112 2 -1 2.002 -0.999001 0.4995005 2 -1 2.001 -0.9995 0.49975012

(X1, Y1) = 2, -1

(X2, Y2) = 2.001, (-2/X2)

(X2, Y2) = (2.001, -0.9995)

= -0.9995-(-1)/2.001-2

=  0.49975012

Comparison Between Tangent & Increment Method:

The table below shows the comparison between the values of gradient obtained from the tangent method and the increment method. Note: the values of the increment method have been rounded to 2.d.p

 Gradient Equation X Tangent Increment 1 -0.95 -1 Y = 1/x -2 -0.25 -0.25 -1 -1.11 -1 2 0.75 0.5 Y = -2/x -2 0.66 0.5 3 0.16 0.18 2 -0.75 -0.75 Y = 3/x -2 -0.75 -0.75 -3 -0.33 -0.33

From the table above we can see that the increment method is much more accurate than the tangent method. Thus I will use the results from the increment method to the generalization of the gradient.

Generalization: Gradient Function For Graphs Of Y = a/x

To make the generalization easier I have drawn a table summarising the gradient found by the increment method at a common point in all the graphs.

 Equation X Gradient Y= 1/x -2 -0.25 Y=-2/x -2 0.5 Y= 3/x -2 -0.75

From the table above we can see that gradient function for the hyperbolas is ‘-a x (1/x) 2

I will now test this equation on the above given points

X = -2 [Y = 1/x]

-a (1/x) 2

= -1(1/-2) 2

= -0.25

X = -2 [Y = -2/x]

-a (1/x) 2

= -(2)(1/-2) 2

= 0.5

X = -2 [Y = 3/x]

-a(1/x)2

= -(3)(1/-2) 2

= -0.75

From the above calculations I conclude that the gradient function of any hyperbola at point ‘x’ is obtained by:

-a (1/x) 2

Graph: Y =X2

The table below shows the increment method performed at the point (1,1).

The values of the gradient have been rounded to 2.d.p

 Y =x2 X1 Y1 X2 Y2 Gradient 1 1 1.005 1.010025 2.00 1 1 1.004 1.008016 2.00 1 1 1.003 1.006009 2.00 1 1 1.002 1.004004 2.00 1 1 1.001 1.002001 2.00

(X1, Y1) = 1,1

(X2, Y2) = 1.001, (X2) 2

(X2, Y2) = (1.001, 1.002001)

= 1.002001-1/1.001-1

=  2.00

We can see from the table above that the gradient at the point (1,1) is 2.

The table below shows the increment method performed at the point (-1,1)

 Y =x2 X1 Y1 X2 Y2 Gradient -1 1 -0.995 0.990025 -1.99 -1 1 -0.996 0.992016 -2.00 -1 1 -0.997 0.994009 -2.00 -1 1 -0.998 0.996004 -2.00 -1 1 -0.999 0.998001 -2.00

(X1, Y1) = -1,1

(X2, Y2) = -0.999, (X2) 2

(X2, Y2) = (-0.999, 0.998001)

= 0.998001-(-1)/ -0.999-1

=  -2.00

From the table above we can see that the value of the gradient at the point X =-1 is 2, since all the increments when rounded to a whole number would give us the same value.

The table below shows the increment method performed at the point (-2,4)

 Y =x2 X1 Y1 X2 Y2 Gradient -2 4 -1.996 3.984016 -4.00 -2 4 -1.997 3.988009 -4.00 -2 4 -1.998 3.992004 -4.00 -2 4 -1.999 3.996001 -4.00

(X1, Y1) = -2,4

(X2, Y2) = -1.999, (X2) 2

(X2, Y2) = (-1.999, 3.996001)

= 3.996001-(-2)/ -1.999-4

=  -4.00

From the table above we can easily see that the gradient at the point X = -2 is –4.

Graph: Y = -2x2

The table below shows the increment method performed at the point (-1, -2)

 Y = -2x2 X1 Y1 X2 Y2 Gradient -1 -2 -0.995 -1.98005 3.99 -1 -2 -0.996 -1.984032 3.99 -1 -2 -0.997 -1.988018 3.99 -1 -2 -0.998 -1.992008 4.00 -1 -2 -0.999 -1.996002 4.00

(X1, Y1) = -1, -2

(X2, Y2) = -0.999, -2(X2) 2

(X2, Y2) = (-0.999, -1.996)

= -1.996-(-2)/ -0.999-(-1)

=  4.00

We can easily see, from the above increment method that the gradient, at the point (-1,2) is 4. There is a slight decrease in the gradient as you go down the X2 values, but this happens only because of the increasing increment. If the values are rounded off to whole numbers, the answer would still be the same, unless of course a very large increment is chosen.

The table below shows the increment method performed at the point (1, -2).

 Y = -2x2 X1 Y1 X2 Y2 Gradient 1 -2 1.005 -2.02005 -4.01 1 -2 1.004 -2.016032 -4.01 1 -2 1.003 -2.012018 -4.01 1 -2 1.002 -2.008008 -4.00 1 -2 1.001 -2.004002 -4.00

(X1, Y1) = 1, -2

(X2, Y2) = 1.001, -2(X2) 2

(X2, Y2) = (1.001, -2.004)

= -2.004-(-2)/ 1.001-1

=  -4.00

From the table above we can see that the gradient at the point (1, -2) is –4.

The table below shows the increment method performed at the point (0,0)

 Y = -2x2 X1 Y1 X2 Y2 Gradient 0 0 0.005 -0.00005 -0.01 0 0 0.004 -0.000032 -0.01 0 0 0.003 -0.000018 -0.01 0 0 0.002 -0.000008 0.00 0 0 0.001 -0.000002 0.00

(X1, Y1) = 0, 0

(X2, Y2) = 0.001, -2(X2) 2

(X2, Y2) = (0.001, -0.0000002)

= -0.0000002-0/ 0.001-0

=  0.00

From the table we see that the gradient at the point (0,0) is 0, however the gradient goes into negative values as the X2 values are increased.

Graph: Y =-0.5x2

The table below shows the increment method performed at the point (-2, -2)

 Y = -0.5x2 X1 Y1 X2 Y2 Gradient -2 -2 -1.995 -1.990013 2.00 -2 -2 -1.996 -1.992008 2.00 -2 -2 -1.997 -1.994005 2.00 -2 -2 -1.998 -1.996002 2.00 -2 -2 -1.999 -1.998001 2.00

(X1, Y1) = -2, -2

(X2, Y2) = -1.999, -0.5(X2) 2

(X2, Y2) = (-1.999, -1.998001)

= -1.998001-(-2)/ -1.999-(-2)

=  2.00

From the table above we can clearly see that at the point (-2, -2) the gradient is 2.

The table below shows the increment method performed at the point (1, -0.5)

 Y = -0.5x2 X1 Y1 X2 Y2 Gradient 1 -0.5 1.005 -0.505013 -1.00 1 -0.5 1.004 -0.504008 -1.00 1 -0.5 1.003 -0.503005 -1.00 1 -0.5 1.002 -0.502002 -1.00 1 -0.5 1.001 -0.501001 -1.00

(X1, Y1) = 1, -0.5

(X2, Y2) = 1.001, -0.5(X2) 2

(X2, Y2) = (1.001, -0.501001)

= -0.501001-(-0.5)/ 1.001-1

=  -1.00

From the table above we can see that at the point (1, -0.5) the gradient is –1.

The table below shows the increment method performed at the point (2, -2)

 Y = -0.5x2 X1 Y1 X2 Y2 Gradient 2 -2 2.005 -2.010013 -2.00 2 -2 2.004 -2.008008 -2.00 2 -2 2.003 -2.006005 -2.00 2 -2 2.002 -2.004002 -2.00 2 -2 2.001 -2.002001 -2.00

(X1, Y1) = 2, -2

(X2, Y2) = 2.001, -0.5(X2) 2

(X2, Y2) = (2.001, -2.002001)

= -2.002001-(-2)/ 2.001-2

= -2

From the table above we can see that the gradient at the point (2, -2) is –2.

Now I have performed the increment method for all the quadratic curves. I will now compare the gradients obtained by the tangent method with the gradients obtained by the increment method.

Comparison Between Tangent & Increment Method:

The table below shows the comparison between the values of the gradients obtained from the tangent method and the increment method:

 Gradient Equation X Tangent Increment -2 -4 -4 Y =x2 -1 -2 -2 1 2 2 -1 5 4 Y =-2x2 0 0 0 1 -5 -4 -2 1.93 2 Y =-0.5x2 1 -1.07 -1 2 -1.8 -2

From the table above we can see that increment method gradients are much more accurate than the tangent method, since in the tangent method a graph is required. The tangent can be drawn wrong if done manually, so I will take the increment method readings to find out the gradient function of the graphs Y =ax2

Generalization: Gradient Function For Graphs Of Y =ax2

To make the Generalization easier I have shown the gradient values from all the graphs at one common point. In this case at the point where X = 1

Equation

X

Y = x2

1

2

Y =-2x2

1

-4

Y =-0.5x2

Conclusion

(y + δy) – y  =  δy

(x + δx) – x  = δx

The symbol δy/ δx is called the derivative or the differential co-efficient of y with respect to x. The procedure used to find δy/ δx is called differentiating y with respect to x. I will now carry out the differentiation method for y = x2 and y = -2x2. I will then aim to find a general formula for δy/ δx when y =xn

δy/ δx Function for x2

Taking P as (x, x2) and a neighbouring point Q as [(x + δx), (x + δx) 2] gives:

δy/ δx Function for -2x2

Y = -2x2

Let δy be the increment in y

Let δx be the increment in x

F (x) = -2x2

F (x + δx) = -2 (x + δx) 2

= 2 (x2 + 2x + δx + (δx) 2

Gradient: F (x + δx) – F (x)

(δx) → 0         δx

=  -2x2 –4x (δx) –2(δx) 2 +2x2

δx

= -4x (δx) - 2 (δx) 2

δx

Limit δx → 0 = -4x - 2 (δx)

δy/ δx = -4x

So when x = 2 δy/ δx = -8

A general formula for δy/ δx when y = xn

The table below shows the gradient function for all the forms of graphs that I have investigated.

 Equation Gradient Y = ax0 0 Y = ax1 a Y = ax2 2ax Y = ax3 3ax2 Y = ax-1 -a x (1/x2)

We can now generalise the gradient functions of all the graphs in the form Y = axn.In general when y = x n where n is any real number:

δy/ δx = nx n-1

Thus the gradient function of any graph in the form Y = axn is:

nxn-1

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

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